My Math Forum Approach reals by 3-smooth numbers

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 October 21st, 2013, 07:43 AM #1 Newbie   Joined: Jun 2012 Posts: 5 Thanks: 0 Approach reals by 3-smooth numbers Hello everyone, I am not a mathematician but I am looking for a theorem. I want to know if it is possible to approximate any real number by the quotient of two 3-smooth numbers. In practice that means: can any real number be approached arbitrary close by $2^m3^n$ where m and n are in $\mathbb{Z}$? Thanks in advance!! Maurice
October 21st, 2013, 08:03 AM   #2
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Re: Approach reals by 3-smooth numbers

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 Originally Posted by Maurice1969 I want to know if it is possible to approximate any real number by the quotient of two 3-smooth numbers.
Well, I don't know of any such theorem, but I doubt whether rationals of the form {1/(2^m * 3^n), 2^m/3^n, 2^m * 3^n} for m,n in N is dense enough in R to make that true.

 October 21st, 2013, 10:15 AM #3 Newbie   Joined: Jun 2012 Posts: 5 Thanks: 0 Re: Approach reals by 3-smooth numbers Thanks for your reply. I doubt it also that every real number can be approached that way, but I really like to know how dense the set is and what kind of set is created by $2^m3^n$ and m,n $\in \mathbb{Z}$, maybe a kind of Cantor set. All info is welcome. Maurice
 October 21st, 2013, 12:57 PM #4 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Approach reals by 3-smooth numbers Well, this isn't a proof, but I believe you can approximate any real number arbitrarily closely with $2^m3^n$ because you can get arbitrarily close to 1. $\log_2 3$ is irrational, and hence you can approximate it arbitrarily closely with rational numbers, which translate into the values of m and n. The closer the ratio m:n is to $\log_2 3$, the closer $2^m3^{-n}$ and $2^{-m}3^n$ are to 1. Remember, you can approximate $\log_2 3$ arbitrarily closely with rational numbers. Now suppose you want to approximate a real number x which is larger than 1. Take an approximation $2^m3^n$ which is sufficiently close to 1 and greater than 1, and then continually multiply it by itself until you get a number which is larger than x, and call the exponent p. Then $(2^m3^n)^p$ will be sufficiently close to x. If the real number x you wish to approximate is less than 1, then simply choose an approximation $2^m3^n$ which is sufficiently close to 1 and less than 1, and continually multiply it by itself until you get a number which is less than x. In particular, the margin of error (percentage wise) in this calculation will be less than or equal to $2^m3^n - 1$, which can be arbitrarily close to 0.
 October 21st, 2013, 01:16 PM #5 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Approach reals by 3-smooth numbers Well obviously the numbers of the form $2^m 3^n$ are all positive and can thus only be dense in the nonnegative real numbers, but that is true. And in any case the full statement on 3-smooth numbers is true. Unfortunately, i don't know of any bigger theorem implying it, so here's the outline of the proof: (It is rather sketchy, but if required I can fill the gaps later) Step 1: It is enough to show: for any $\varepsilon > 0$ there are integral m, n such that $1 < 2^m 3^n < 1 + \varepsilon$. In fact, assuming the assertion: For every real r > 1 and every $\varepsilon > 0$ take f = 2^m 3^m to be a number as described in the assertion, i.e. 1 < f < 1 + e. Then there is an integer k such that $f \leq f^k < f(1+ \varepsilon )$. Letting epsilon go to 0 we can thus approximate r. The proof for 0< r < 1 is similar. Step 2: It is enough to show: for any $\varepsilon > 0$ there are integral m, n such that $0 < m * log( 2 ) + n * log( 3 ) < \varepsilon$. To prove the assertion of step 1 for any given $\varepsilon > 0$ assume the assertion of step 2 is valid $\varepsilon ' = log( 1 + \varepsilon )$. Then the same pair (m,n) will satisfy the conditions of step 1. Step 3: The assertion of step 2 is well known to be true for any two nonzero reals a,b in stead of log( 2) and log (3) such that the quotient a/b is irrational. There is a nice proof using topological group theory, but i don't know a reference at the moment. maybe someone else knows one.
 October 21st, 2013, 04:04 PM #6 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Approach reals by 3-smooth numbers Hi, Yes any number can be approached in this way. As Peter as stated this is equivalent to ask if $m\log2+n\log3$ is dense in R. Note that this space forms a group, which is a subgroup of R. It is easy to show that any subgroups of R are either of the form $aZ$ with a is a real number or are dense in R. Obviously there is no such a in our case, the group has no smallest elements therefor this group is dense in R and any real number can be approached.
October 21st, 2013, 11:12 PM   #7
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Re: Approach reals by 3-smooth numbers

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 Originally Posted by Peter Step 2: It is enough to show: for any $\varepsilon > 0$ there are integral m, n such that $0 < m * log( 2 ) + n * log( 3 ) < \varepsilon$. .
Yes, I thought of that after I have questioned it's density, that the density of n log 2 + m log 3 might suffice, but haven't had the time to get through a formal outline.

So, definitely, nice thinking and nice proof, Peter.

 October 22nd, 2013, 06:00 AM #8 Newbie   Joined: Jun 2012 Posts: 5 Thanks: 0 Re: Approach reals by 3-smooth numbers Thank you all for the replies! It helped me a lot. Is it possible to say now that every positive real can be approached by $3^m / 2^n$ with m and n in $\mathbb{N}$ since it can be very close to one. Thanks again!
October 22nd, 2013, 07:31 AM   #9
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Re: Approach reals by 3-smooth numbers

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 Originally Posted by Maurice1969 Is it possible to say now that every positive real can be approached by $3^m / 2^n$ with m and n in $\mathbb{N}$ since it can be very close to one.
That'd be false. For example, take any real in the interval [0, 1].

October 22nd, 2013, 01:34 PM   #10
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Re: Approach reals by 3-smooth numbers

Quote:
 Originally Posted by Maurice1969 Thank you all for the replies! It helped me a lot. Is it possible to say now that every positive real can be approached by $3^m / 2^n$ with m and n in $\mathbb{N}$ since it can be very close to one. Thanks again!
My understanding is that this statement is true, but the reason it is true is thus: $\frac{3^m}{2^n}$ can be arbitrarily close to 1 (very close is not good enough) because $\log_2 3$ is irrational (which is always true for two numbers which are relatively prime). Being arbitrarily close means you can be within any positive distance from 1 on the number line. Note that this statement is also true for the fraction $\frac{2^m}{3^n}$, where m and n are natural numbers. It is not true for $2^m3^n$ where m and n are natural numbers, since the only numbers obtainable in this fashion are integers, which cannot be arbitrarily close to 1.

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